Let $\tau$ be a linear operator on $\mathbb{C}^n$ and let $\lambda_{1},...,\lambda_{n}$ be the eigenvalues of $\tau$, each one written a number of times equal to its algebraic multiplicity. I should show that:

$\sum_{i}|\lambda_{i}|^2\leq tr(\tau^*\tau)$

Also, one should show that the equality holds iff $\tau$ is normal.

First I felt that this might use singular values, but I have no success with this. My idea then was that Cauchy-Schwarz may be useful. (I work with matrices, this is clearly not a restriction to the problem.) So I defined the inner-product $\langle A,B\rangle=tr(B^*A)$, which I know to be acceptable. Elementary operations on Cauchy-Schwarz inequality

2 Answers
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We have a Schur decomposition $\tau=u^*\sigma u$ with $u$ unitary and $\sigma$ an upper triangular matrix. Since both sides of the inequality take the same value for $\tau$ and for $\sigma$ we may assume that $\tau$ is in fact itself upper triangular. Now the inequality is obvious, because the right hand side is a sum of the left hand side and some other, non-negative terms.

I fail to follow your argument. For diagonal matrices, the inequality in the question is an equality. Why and how would that extend as a (proper) inequality to more general matrices?
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Martin ArgeramiSep 6 '12 at 1:46

As an example of where the inequality is strict, consider the matrix representation of the right shift operator $R$, $e_1 \rightarrow e_2 \rightarrow e_3 \rightarrow 0$ and it's adjoint the left shift operator $L=R^*$, $e_3 \rightarrow e_2 \rightarrow e_1 \rightarrow 0$. The composition $RL$ acts as the identity on $e_1$ and $e_2$, and sends $e_3$ to zero, so it has trace 2, whereas there are no eigenvalues of $R$ or $L$.
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NickSep 6 '12 at 6:03

N.B. I have just edited my answer, and the two comments aobve refer to the old version.
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Mariano Suárez-Alvarez♦Sep 6 '12 at 6:27

It wouldn't surprise me if I'm missing something, but I couldn't think of a straightforward proof. But this inequality is a particular case of Weyl's Majorant Theorem (see Theorem II.3.6 in Bhatia's Matrix Analysis, or there are surely many other references).

You are welcome. I couldn't address the implication "left implies right" with this idea. But it does follow easily from Mariano's argument (the case of equality forces $\sigma$ to be diagonal).
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Martin ArgeramiSep 6 '12 at 14:17